Forward product change repair keeps the joint increment

Forward product change repair starts by retaining the change shared by both factors. In this invented exact calculation, a product moves from 5 times 7 to 6 times 9. Its change is 54 minus 35, or 19. A shortcut that adds only 5 times 2 and 7 times 1 returns 17. The missing two is the product of the two factor changes, one and two.

Define the two states before calculating

Let f and g be real scalar quantities recorded at two positions, n and n + 1. Write f for the first value and f + Δf for the next value. Likewise, write g and g + Δg for the other quantity. The symbol Δ means next value minus first value. It permits a positive, negative or zero change; it does not mean a percentage.

The requested output is the next product minus the first product: (f + Δf)(g + Δg) − fg. This is an exact finite change between two specified states. It is not a rate estimated from a device, a financial return or a claim that either product listing performs a calculation. There is no rounding, measurement error or unstated intermediate observation in the model.

For the main example, stipulate f(n) = n + 2 and g(n) = 2n + 1, then use n = 3. These formulas are supplied inputs. They give first values f(3) = 5 and g(3) = 7, and next values f(4) = 6 and g(4) = 9. Thus Δf = 1 and Δg = 2. Keep those four values together rather than borrowing a next value from a different case.

Check the requested change directly

The first product is 5 × 7 = 35. The next product is 6 × 9 = 54. Subtracting in the stated direction gives 54 − 35 = 19. This direct calculation is an independent check on any formula proposed for the change. The number 54 alone is the next product; it is not the change, because the first product must still be subtracted.

The proposed shortcut uses the original factors: fΔg + gΔf. Substitution gives 5 × 2 + 7 × 1 = 10 + 7 = 17. Both multiplications and the addition are correct. The error is in treating their sum as the entire requested change. Preserve 17 beside the shortcut that produced it, then identify what that expression leaves out.

Expand both changed factors

Distribute the two factors in the next product. Choosing f from the first factor and g from the second gives fg. Choosing f and Δg gives fΔg. Choosing Δf and g gives gΔf, since these are ordinary real scalars. Choosing both changes gives ΔfΔg. Therefore the next product is fg + fΔg + gΔf + ΔfΔg.

Now subtract the original product fg. That removes the unchanged contribution and leaves the exact identity Δ(fg) = fΔg + gΔf + ΔfΔg. Every term uses the same two input states. The final term belongs to the expansion even when both individual changes seem small. An exact calculation cannot discard it merely because it is smaller than the other terms.

In the worked example the three contributions are ten, seven and two. Their sum is nineteen, matching the direct subtraction. The difference between the actual result and the shortcut is exactly ΔfΔg. This identifies both the missing amount and its origin. Changing an original factor to make seventeen agree would alter the problem rather than repair the proposed calculation.

Use a shifted factor for a correct two-term rule

A two-term identity is possible if one factor comes from the next state. Group fΔg with ΔfΔg to obtain (f + Δf)Δg. The full change then becomes (f + Δf)Δg + gΔf. In the original index notation this is f(n + 1)Δg + g(n)Δf. The shift in the first factor carries the joint contribution inside that term.

For our inputs, the shifted calculation is 6 × 2 + 7 × 1 = 12 + 7 = 19. Alternatively, group the joint term with gΔf and use fΔg + (g + Δg)Δf, giving 5 × 2 + 9 × 1 = 19. Both versions work. Keep their state labels explicit: one original factor and one next factor appear in each correct version.

Using both next factors would be another shortcut, and generally another error. Here 6 × 2 + 9 × 1 gives 21. Expanding that expression shows why: it includes the joint increment twice, while the actual product change includes it once. The choice is not simply whether newer values are better. The identity specifies which value belongs in each position.

State exactly when the original shortcut agrees

Subtract the original two-term shortcut from the full identity. The residual is ΔfΔg. Over the real numbers, that residual is zero exactly when at least one of Δf and Δg is zero. Thus the original-factor shortcut works when one factor does not change, including the case where neither changes. This is a condition on the changes, rather than on whether an original factor happens to be zero.

For example, retain f = 5, Δf = 1 and g = 7, but set Δg = 0. The product moves from 35 to 42, a change of seven. The shortcut gives 5 × 0 + 7 × 1 = 7, and the joint term is zero. This successful control does not establish the shortcut for cases where both factors change. Its success has a precise stated reason.

Check a decreasing factor without losing its sign

Use a separate input record in which f still goes from five to six, while g goes from seven to five. Now Δf = 1 and Δg = −2. The product changes from 35 to 30, so the requested change is −5. A negative answer is appropriate: the next product is smaller. Replacing the decrease with a positive magnitude would change the defined operation.

The original-factor shortcut gives 5 × (−2) + 7 × 1 = −3. The joint term is 1 × (−2) = −2, bringing the full answer to −5. Here omitting the joint term makes the answer too high, rather than too low as in the main example. The error's direction depends on the sign of the product of changes; it is not always an underestimate.

Keep the square case as a special example

If the two factors are the same quantity, replace g with f and Δg with Δf. The exact identity becomes Δ(f²) = 2fΔf + (Δf)². For a change from five to six, the square changes from twenty-five to thirty-six, a difference of eleven. Twice the original five times the change one gives ten; the remaining square of the change supplies one.

This example uses commutative scalar multiplication throughout. No reversal of multiplication order explains its missing term. Even when the two individual contributions combine into 2fΔf, the joint square remains. The general product identity supplies the square case by substitution, so it should not be used as a reason to omit a term from the more general two-factor calculation.

Test a fractional step exactly

Return to the supplied functions at index three, but compare them with index 3 + h for a declared nonzero real step h. Their changes are h and 2h. Direct substitution into the product-change identity gives 5(2h) + 7h + 2h² = 17h + 2h². At h = 1/2, the next factors are 11/2 and 8, whose product is 44.

The direct change is therefore 44 − 35 = 9. The original-factor contributions total 17/2, and the joint contribution is 1/2, restoring nine. If the requested output were change divided by step, division would give 17 + 2h, which is eighteen at this half-step. The finite-step quotient is a separate requested quantity. Setting h to zero inside its division is not permitted.

Verify the entire integer-index family

The known product of the main functions is (n + 2)(2n + 1) = 2n² + 5n + 2. At the next integer index it becomes 2(n + 1)² + 5(n + 1) + 2. Expanding and subtracting the first expression gives 4n + 7. At n = 3 this is nineteen, agreeing with the original two-state calculation.

The shortcut instead gives (n + 2) × 2 + (2n + 1) × 1 = 4n + 5. Its residual is always two for this family because the two one-step changes remain one and two. An exact control also checked all 2,401 choices of f, g, Δf and Δg from the integers −3 through 3. Every choice satisfied both corrected identities; the algebra above establishes them beyond that finite control set.

Read the added corner in the positive example

For a conceptual picture of the main positive inputs, draw a rectangle with side values five and seven. Extend those sides by one and two. The added portions consist of a five-by-two strip, a seven-by-one strip and a one-by-two corner. Their areas are ten, seven and two. The corner represents the joint increment omitted by the original shortcut.

This drawing interprets the positive scalar example rather than measuring an actual object. The decreasing-factor control is still settled by the signed algebra, without pretending that a negative change is an ordinary added patch. A schematic cover can highlight the corner, but the four input values and exact expansion remain the proof of the numerical result.

Keep the native accounts within their evidence

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The GLOWNEON owl bookend account shows flat owl-and-book artwork with example lettering around upright books. The real books and plant are staging; drawn books belong to the decoration. Its left-side, right-side and full-set choices do not establish a delivered pair. Maker authenticity, measurements, material and performance remain unverified.

Distinguish neighboring mathematical repairs

The finite differences worksheet recovers unknown coefficients within a supplied quadratic family. Here both functions are known at the outset. The repair concerns how an exact difference acts on their product, rather than using observations to fit a function or extend an unspecified sequence.

The matrix square expansion repair retains two ordered cross products because their matrix values can differ. This example uses commuting real scalars and two different states. Its joint increment survives ordinary scalar expansion; the correct two-term alternative shifts one factor to the next state.

Retain the original and next factors, their signed changes, the two products and the requested subtraction direction. For the main record, keep nineteen beside the direct check, seventeen beside the failed original-factor shortcut, and two beside the omitted joint increment. The repaired identity then explains the agreement without changing an input or treating a favorable special case as a general rule.

Keep the joint increment with the two individual changes.

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